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  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Chapter 5Joint Probability Distributions

    Applied Statistics and Probability for

    Engineers

    Sixth Edition

    Douglas C. Montgomery George C. Runger

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Chapter 5 Title and Outline2

    5Joint Probability

    Distributions

    5-1 Two or More Random Variables

    5-1.1 Joint Probability Distributions

    5-1.2 Marginal Probability Distributions

    5-1.3 Conditional Probability Distributions

    5-1.4 Independence

    5-1.5 More Than Two Random Variables

    5-2 Covariance and Correlation

    5-3 Common Joint Distributions

    5-3.1 Multinomial Probability Distribution

    5-3.2 Bivariate Normal Distribution

    5-4 Linear Functions of Random Variables

    5-5 General Functions of Random Variables

    5-6 Moment Generating Functions

    CHAPTER OUTLINE

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Learning Objectives for Chapter 5

    After careful study of this chapter, you should be able to do the

    following:1. Use joint probability mass functions and joint probability density

    functions to calculate probabilities.

    2. Calculate marginal and conditional probability distributions from joint probability distributions.

    3. Interpret and calculate covariances and correlations between random variables.

    4. Use the multinomial distribution to determine probabilities.

    5. Properties of a bivariate normal distribution and to draw contour plots for the probability density function.

    6. Calculate means and variances for linear combinations of random variables, and calculate probabilities for linear combinations of normally distributed random variables.

    7. Determine the distribution of a general function of a random variable.

    8. Calculate moment generating functions and use them to determine moments and distributions

    3Chapter 5 Learning Objectives

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Joint Probability Mass Function

    Sec 5-1.1 Joint Probability Distributions 4

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Joint Probability Density Function

    Sec 5-1.1 Joint Probability Distributions 5

    Figure 5-2 Joint probability

    density function for the random

    variables X and Y. Probability

    that (X, Y) is in the region R is

    determined by the volume of

    fXY(x,y) over the region R.

    The joint probability density function for the continuous

    random variables X and Y, denotes as fXY(x,y), satisfies the

    following properties:

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-2: Server Access Time-1Let the random variable X denote the time until a computer

    server connects to your machine (in milliseconds), and let Y

    denote the time until the server authorizes you as a valid user (in

    milliseconds). X and Y measure the wait from a common

    starting point (x < y). The joint probability density function for X

    and Y is

    Sec 5-1.1 Joint Probability Distributions 6

    Figure 5-4 The joint probability

    density function of X and Y is

    nonzero over the shaded region

    where x < y.

    0.001 0.002 6, for 0 and 6 10x yXYf x y ke x y k

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-2: Server Access Time-2

    The region with nonzero probability is shaded in

    Fig. 5-4. We verify that it integrates to 1 as follows:

    Sec 5-1.1 Joint Probability Distributions 7

    0.001 0.002 0.002 0.001

    0 0 0 0

    0.0020.001 0.003

    0 0

    ,

    0.0030.002

    10.003 1

    0.003

    x y y x

    XY

    xx x

    f x y dydx ke dy dx k e dy e dx

    ek e dx e dx

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-2: Server Access Time-3

    Now calculate a probability:

    Sec 5-1.1 Joint Probability Distributions 8

    Figure 5-5 Region of

    integration for the probability

    that X < 1000 and Y < 2000

    is darkly shaded.

    1000 2000

    0

    1000 2000

    0.002 0.001

    0

    1000 0.002 40.001

    0

    1000

    0.003 4 0.001

    0

    3 14

    1000, 2000 ,

    0.002

    0.003

    1 10.003

    0.003 0.001

    XY

    x

    y x

    x

    xx

    x x

    P X Y f x y dydx

    k e dy e dx

    e ek e dx

    e e e dx

    e ee

    0.003 316.738 11.578 0.915

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Marginal Probability Distributions (discrete)

    The marginal probability distribution for X is found by summing the probabilities

    in each column whereas the marginal probability distribution for Y is found by

    summing the probabilities in each row.

    Sec 5-1.2 Marginal Probability Distributions 9

    X

    y

    Y

    x

    f x f xy

    f y f xy

    1 2 3 f (y )

    1 0.01 0.02 0.25 0.28

    2 0.02 0.03 0.20 0.25

    3 0.02 0.10 0.05 0.17

    4 0.15 0.10 0.05 0.30

    f (x ) 0.20 0.25 0.55 1.00

    y = Response

    time(nearest

    second)

    x = Number of Bars of

    Signal Strength

    Marginal probability distributions of X and Y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Marginal Probability Density Function (continuous)

    If the joint probability density function of

    random variables X and Y is fXY(x,y), the

    marginal probability density functions of X

    and Y are:

    Sec 5-1.2 Marginal Probability Distributions 10

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-4: Server Access Time-1

    For the random variables that

    denotes times in Example 5-2,

    find the probability that Y

    exceeds 2000 milliseconds.

    Integrate the joint PDF directly

    using the picture to determine

    the limits.

    Sec 5-1.2 Marginal Probability Distributions 11

    2000

    0 2000 2000

    2000 , ,

    Dark region left dark region right dark region

    XY XY

    x

    P Y f x y dy dx f x y dy dx

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-4: Server Access Time-2

    Alternatively, find the marginal PDF and then

    integrate that to find the desired probability.

    Sec 5-1.2 Marginal Probability Distributions 12

    0.001 0.002

    0

    0.002 0.001

    0

    0.0010.002

    0

    0.0010.002

    3 0.002 0.001

    0.001

    1

    0.001

    6 10 1 for 0

    y

    x y

    Y

    y

    y x

    yx

    y

    yy

    y y

    f y ke dx

    ke e dx

    eke

    eke

    e e y

    2000

    3 0.002 0.001

    2000

    0.002 0.0033

    2000 2000

    4 63

    2000

    6 10 1

    6 100.002 0.003

    6 10 0.050.002 0.003

    Y

    y y

    y y

    P Y f y dy

    e e dy

    e e

    e e

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean & Variance of a Marginal Distribution

    E(X) and V(X) can be obtained by first calculating the marginal

    probability distribution of X and then determining E(X) and V(X) by

    the usual method.

    Sec 5-1.2 Marginal Probability Distributions 13

    2 2

    2 2

    X

    R

    X X

    R

    Y

    R

    Y Y

    R

    E X x f x

    V X x f x

    E Y y f y

    V Y y f y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean & Variance for Example 5-1

    Sec 5-1.2 Marginal Probability Distributions 14

    1 2 3

    1 0.01 0.02 0.25 0.28 0.28 0.28

    2 0.02 0.03 0.20 0.25 0.50 1.00

    3 0.02 0.10 0.05 0.17 0.51 1.53

    4 0.15 0.10 0.05 0.30 1.20 4.80

    f (x ) 0.20 0.25 0.55 1.00 2.49 7.61

    x *f (x ) 0.20 0.50 1.65 2.35

    x 2*f (x ) 0.20 1.00 4.95 6.15

    y 2*f (y )

    x = Number of Bars

    of Signal Strength

    y = Response

    time(nearest

    second)

    f (y ) y *f (y )

    E(X) = 2.35 V(X) = 6.15 – 2.352 = 6.15 – 5.52 = 0.6275

    E(Y) = 2.49 V(Y) = 7.61 – 2.492 = 7.61 – 16.20 = 1.4099

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Conditional Probability Density Function

    Sec 5-1.3 Conditional Probability Distributions 15

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-6: Conditional Probability-1From Example 5-2, determine the conditional PDF for Y given X=x.

    Sec 5-1.3 Conditional Probability Distributions 16

    0.001 0.002

    0.0020.001

    0.0020.001

    0.003

    0.001 0.002

    0.003

    0.002 0.002

    0.002

    0.002

    0.003 for 0

    ,

    ( ) 0.003

    0.002 for 0

    x y

    X

    x

    yx

    x

    x

    x

    x yXY

    Y x x

    X

    x y

    f x k e dy

    eke

    eke

    e x

    f x y kef y

    f x e

    e x and x y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-6: Conditional Probability-2

    Now find the probability that Y exceeds 2000 given that X=1500:

    Sec 5-1.3 Conditional Probability Distributions 17

    1500

    2000

    0.002 1500 0.002

    2000

    0.0023

    2000

    43 1

    2000 1500

    0.002

    0.0020.002

    0.002 0.3680.002

    Y

    y

    y

    P Y X

    f y dy

    e

    ee

    ee e

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean & Variance of Conditional Random Variables

    • The conditional mean of Y given X = x,

    denoted as E(Y|x) or μY|x is

    • The conditional variance of Y given X = x,

    denoted as V(Y|x) or σ2Y|x is

    Sec 5-1.3 Conditional Probability Distributions 18

    Y xy

    E Y x y f y

    2

    2 2

    Y x Y x Y x Y x

    y y

    V Y x y f y y f y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-8: Conditional Mean And Variance

    From Example 5-2 & 5-6, what is the conditional mean for

    Y given that x = 1500?

    Sec 5-1.3 Conditional Probability Distributions 19

    0.002 1500 0.002 3 0.002

    1500 1500

    0.002 0.0023

    15001500

    0.0023 3

    1500

    3

    1500 0.002 0.002

    0.0020.002 0.002

    15000.002

    0.002 0.002 0.002

    0.002

    y y

    y y

    y

    E Y X y e dy e y e dy

    e ee y dy

    ee e

    e

    33

    33

    1500

    0.002 0.002 0.002

    0.002 2000 20000.002

    ee

    ee

    If the connect time is 1500 ms, then the expected time to be authorized is 2000 ms.

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-9For the discrete random variables in Exercise 5-1,

    what is the conditional mean of Y given X=1?

    Sec 5-1.3 Conditional Probability Distributions 20

    1 2 3

    1 0.01 0.02 0.25 0.28

    2 0.02 0.03 0.20 0.25

    3 0.02 0.10 0.05 0.17

    4 0.15 0.10 0.05 0.30

    f (x ) 0.20 0.25 0.55 y*f(y|x=1) y2*f(y|x=1)

    1 0.050 0.080 0.455 0.05 0.05

    2 0.100 0.120 0.364 0.20 0.40

    3 0.100 0.400 0.091 0.30 0.90

    4 0.750 0.400 0.091 3.00 12.00

    Sum of f(y|x) 1.000 1.000 1.000 3.55 13.35

    12.6025

    0.7475

    f (y )y = Response

    time(nearest

    second)

    x = Number of Bars

    of Signal Strength

    The mean number of attempts given one bar is 3.55 with variance of 0.7475.

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Independent Random Variables

    For random variables X and Y, if any one of the

    following properties is true, the others are also true.

    Then X and Y are independent.

    Sec 5-1.4 Independence 21

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-11: Independent Random Variables

    • Suppose the Example 5-2 is modified such that the joint

    PDF is:

    • Are X and Y independent?

    • Find the probability

    Sec 5-1.4 Independence 22

    6 0.001 0.002, 2 10 for 0 and 0.x yXYf x y e x y

    6 0.001 0.002

    0

    0.001

    2 10

    0.001 for 0

    x y

    X

    x

    f x e dy

    e x

    6 0.001 0.002

    0

    0.002

    2 10

    0.002 for y > 0

    x y

    Y

    y

    f y e dx

    e

    1 21000, 1000 1000 1000

    1 0.318

    P X Y P X P Y

    e e

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Joint Probability Density Function

    Sec 5-1.5 More Than Two Random Variables 23

    The joint probability density function for the continuous

    random variables X1, X2, X3, …Xp, denoted as

    satisfies the following properties: 1 2 ... 1 2

    , ,...,pX X X p

    f x x x

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-14: Component Lifetimes

    In an electronic assembly, let X1, X2, X3, X4 denote

    the lifetimes of 4 components in hours. The joint

    PDF is:

    What is the probability that the device operates

    more than 1000 hours?

    The joint PDF is a product of exponential PDFs.

    P(X1 > 1000, X2 > 1000, X3 > 1000, X4 > 1000)

    = e-1-2-1.5-3 = e-7.5 = 0.00055

    Sec 5-1.5 More Than Two Random Variables 24

    1 2 3 41 2 3 4

    0.001 0.002 0.0015 0.00312

    1 2 3 4, , , 9 10 for x 0x x x x

    X X X X if x x x x e

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Marginal Probability Density Function

    Sec 5-1.5 More Than Two Random Variables 25

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean & Variance of a Joint Distribution

    The mean and variance of Xi can be

    determined from either the marginal PDF, or

    the joint PDF as follows:

    Sec 5-1.5 More Than Two Random Variables 26

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-16Points that have positive probability in the

    joint probability distribution of three random

    variables X1 , X2 , X3 are shown in Figure.

    Suppose the 10 points are equally likely

    with probability 0.1 each. The range is the

    non-negative integers with x1+x2+x3 = 3

    List the marginal PDF of X2

    Sec 5-1.5 More Than Two Random Variables 27

    P (X2 = 0) = (3,0,0) + (0,0,3) + (1,0,2) + (2,0,1) = 0.4

    P (X2 = 1) = (2,1,0) + (0,1,2) + (1,1,1) = 0.3

    P (X2 = 2) = (1,2,0) + (0,2,1) = 0.2

    P (X2 = 3) = (0,3,0) = 0.1

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    321 xxxf

    Also, E(x2) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Distribution of a Subset of Random Variables

    Sec 5-1.5 More Than Two Random Variables 28

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Conditional Probability Distributions

    • Conditional probability distributions can be

    developed for multiple random variables by

    extension of the ideas used for two random

    variables.

    • Suppose p = 5 and we wish to find the

    distribution conditional on X4 and X5.

    Sec 5-1.5 More Than Two Random Variables 29

    1 2 3 4 5

    1 2 3 4 5

    4 5

    4 5

    1 2 3 4 5

    1 2 3

    4 5

    4 5

    , , , ,, ,

    ,

    for , 0.

    X X X X X

    X X X X X

    X X

    X X

    f x x x x xf x x x

    f x x

    f x x

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Independence with Multiple Variables

    The concept of independence can be extended to

    multiple variables.

    Sec 5-1.5 More Than Two Random Variables 30

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-18: Layer ThicknessSuppose X1,X2, and X3 represent the thickness in μm of a

    substrate, an active layer and a coating layer of a chemical

    product. Assume that these variables are independent and

    normally distributed with parameters and specified limits as

    tabled.

    Sec 5-1.5 More Than Two Random Variables 31

    X 1 X 2 X 3

    Mean (μ) 10,000 1,000 80

    Std dev (σ) 250 20 4

    Lower limit 9,200 950 75

    Upper limit 10,800 1,050 85

    P(in limits) 0.99863 0.98758 0.78870

    0.77783P(all in limits) =

    Normal

    Random VariablesParameters

    and specified

    limits

    What proportion of the product

    meets all specifications?

    Answer: 0.7783, 3 layer product.

    Which one of the three

    thicknesses has the least

    probability of meeting specs?

    Answer: Layer 3 has least prob.

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Covariance

    • Covariance is a measure of the relationship between two random variables.

    • First, we need to describe the expected value of a function of two random variables. Let h(X, Y) denote the function of interest.

    Sec 5-2 Covariance & Correlation 32

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-19: Expected Value of a Function of Two Random Variables

    Sec 5-2 Covariance & Correlation 33

    For the joint probability distribution of the two random variables in

    Example 5-1, calculate E [(X-μX)(Y-μY)].

    The result is obtained by multiplying x - μX times y - μY, times fxy(X,Y)

    for each point in the range of (X,Y). First, μX and μy were determined

    previously from the marginal distributions for X and Y:

    μX = 2.35 and μy = 2.49

    Therefore,

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Covariance Defined

    Sec 5-2 Covariance & Correlation 34

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Correlation (ρ = rho)

    Sec 5-2 Covariance & Correlation 35

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-21: Covariance & Correlation

    Determine the covariance

    and correlation to the figure

    below.

    Sec 5-2 Covariance & Correlation 36

    x y f(x, y) x-μX y-μY Prod

    0 0 0.2 -1.8 -1.2 0.42

    1 1 0.1 -0.8 -0.2 0.01

    1 2 0.1 -0.8 0.8 -0.07

    2 1 0.1 0.2 -0.2 0.00

    2 2 0.1 0.2 0.8 0.02

    3 3 0.4 1.2 1.8 0.88

    0 0.2 1.260

    1 0.2 0.926

    2 0.2

    3 0.4

    0 0.2

    1 0.2

    2 0.2

    3 0.4

    μX = 1.8

    μY = 1.8

    σX = 1.1662

    σY = 1.1662StD

    ev

    correlation =

    Mar

    gin

    al

    covariance =

    Mea

    n

    Note the strong

    positive correlation.

    Join

    t

    Figure 5-13 Discrete joint

    distribution, f(x, y).

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Independence Implies ρ = 0

    • If X and Y are independent random

    variables,

    σXY = ρXY = 0

    • ρXY = 0 is necessary, but not a sufficient

    condition for independence.

    Sec 5-2 Covariance & Correlation 37

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-23: Independence Implies Zero Covariance

    Sec 5-2 Covariance & Correlation 38

    Let 16 for 0 2 and 0 4

    Show that 0

    XY

    XY

    f xy x y x y

    E XY E X E Y

    Figure 5-15 A planar

    joint distribution.

    4 2

    2 2

    0 0

    24 32

    0 0

    4

    2

    0

    43

    0

    1

    16

    1

    16 3

    1 8

    16 3

    1 1 64 32

    6 3 6 3 9

    .

    32 4 80

    9 3 3

    XY

    E XY x y dx dy

    xy dy

    y dy

    y

    E XY E X E Y

    4 2

    2

    0 0

    24 3

    0 0

    42

    0

    4 2

    2

    0 0

    24 22

    0 0

    43

    0

    1

    16

    1

    16 3

    1 8 1 16 4

    16 2 3 6 2 3

    1

    16

    1

    16 2

    2 1 64 8

    16 3 8 3 3

    E X x ydx dy

    xy dy

    y

    E Y xy dx dy

    xy dy

    y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Multinomial Probability Distribution

    • Suppose a random experiment consists of a series of n trials. Assume that:1) The outcome of each trial can be classifies into one of k

    classes.

    2) The probability of a trial resulting in one of the k outcomes is constant, and equal to p1, p2, …, pk.

    3) The trials are independent.

    • The random variables X1, X2,…, Xk denote the number of outcomes in each class and have a multinomial distribution and probability mass function:

    Sec 5-3.1 Multinomial Probability Distribution 39

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-25: Digital Channel

    Of the 20 bits received over a digital channel, 14 are of excellent

    quality, 3 are good, 2 are fair, 1 is poor. The sequence received was

    EEEEEEEEEEEEEEGGGFFP. Let the random variables X1 , X2 , X3,

    and X4 denote the number of bits that are E, G, F , and P, respectively,

    in a transmission of 20 bits. What is the probability that 12 bits are E, 6

    bits are G, 2 are F, and 0 are P?

    Sec 5-3.1 Multinomial Probability Distribution 40

    12 6 2 01 2 3 420!

    12,X 6,X 2,X 0 0.6 0.3 0.08 0.02 0.035812!6!2!0!

    P X

    0.03582 = (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2

    Using Excel

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Multinomial Mean and Variance

    The marginal distributions of the multinomial

    are binomial.

    If X1, X2,…, Xk have a multinomial distribution,

    the marginal probability distributions of Xi is

    binomial with:

    E(Xi) = npi and V(Xi) = npi(1-pi)

    Sec 5-3.1 Multinomial Probability Distribution 41

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Bivariate Normal Probability Density Function

    Sec 5-3.2 Bivariate Normal Distribution 42

    2

    2 2

    2 22

    1, ; , , , ,

    2 1

    21where

    2 1

    for and .

    u

    XY X X Y Y

    X Y

    X X Y Y

    X X Y Y

    f x y e

    x x y yu

    x y

    0, ,Parameter limits: 1 1

    0, ,

    x x

    y y

    The probability density function of a bivariate normal distribution is

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Marginal Distributions of the Bivariate Normal Random Variables

    If X and Y have a bivariate normal distribution with

    joint probability density function

    fXY(x,y;σX,σY,μX,μY,ρ)

    the marginal probability distributions of X and Y

    are normal with means μX and μY and standard

    deviations σX and σY, respectively.

    Sec 5-3.2 Bivariate Normal Distribution 43

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Conditional Distribution of Bivariate Normal Random Variables

    If X and Y have a bivariate normal distribution with

    joint probability density fXY(x,y;σX,σY,μX,μY,ρ), the

    conditional probability distribution of Y given X = x is

    normal with mean and variance as follows:

    Sec 5-3.2 Bivariate Normal Distribution 44

    2 2 21

    YY XY x

    X

    YY x

    x

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Correlation of Bivariate Normal Random Variables

    If X and Y have a bivariate normal

    distribution with joint probability density

    function fXY(x,y;σX,σY,μX,μY,ρ), the

    correlation between X and Y is ρ.

    Sec 5-3.2 Bivariate Normal Distribution 45

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Bivariate Normal Correlation and Independence

    • In general, zero correlation does not imply independence.

    • But in the special case that X and Y have a bivariate normal distribution, if ρ = 0, then X and Y are independent.

    If X and Y have a bivariate normal

    distribution with ρ=0, X and Y are

    independent.

    Sec 5-3.2 Bivariate Normal Distribution 46

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    Linear Functions of Random Variables

    • A function of random variables is itself a random variable.

    • A function of random variables can be formed by either linear or nonlinear relationships. We limit our discussion here to linear functions.

    • Given random variables X1, X2,…,Xp and constants c1, c2, …, cp

    Y= c1X1 + c2X2 + … + cpXpis a linear combination of X1, X2,…,Xp.

    Sec 5-4 Linear Functions of Random Variables 47

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean and Variance of a Linear Function

    If X1, X2,…,Xp are random variables, and Y= c1X1 + c2X2 +

    … + cpXp , then

    Sec 5-4 Linear Functions of Random Variables 48

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-31: Error Propagation

    A semiconductor product consists of three layers.

    The variances of the thickness of each layer is 25,

    40 and 30 nm. What is the variance of the finished

    product?

    Answer:

    Sec 5-4 Linear Functions of Random Variables 49

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Mean and Variance of an Average

    Sec 5-4 Linear Functions of Random Variables 50

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Reproductive Property of the Normal Distribution

    Sec 5-4 Linear Functions of Random Variables 51

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    Example 5-32: Linear Function of Independent Normal Random variables

    Let the random variables X1 and X2 denote

    the length and width of a manufactured

    part. Their parameters are shown in the

    table. What is the probability that the

    perimeter exceeds 14.5 cm?

    Sec 5-4 Linear Functions of Random Variables 52

    X 1 X 2

    Mean 2 5

    Std Dev 0.1 0.2

    Parameters of

    1 2

    1 2

    2 22 2

    1 2

    Let 2 2 perimeter

    2 2 2 2 2 5 14 cm

    2 2 4 0.1 4 0.2 0.04 0.16 0.20

    0.20 0.4472 cm

    14.5 1414.5 1 1 1.1180 0.1318

    .4472

    Y X X

    E Y E X E X

    V Y V X V X

    SD Y

    P Y

    0.1318 = 1 - NORMDIST(14.5, 14, SQRT(0.2), TRUE)

    Using Excel

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    General Function of a Discrete Random Variable

    Suppose that X is a discrete random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of Xand Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability mass function of the random variable Y is

    fY(y) = fX[u(y)]

    Sec 5-5 General Functions of Random Variables 53

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-34: Function of a Discrete Random Variable

    Let X be a geometric random variable with probability

    distribution

    fX(x) = p(1-p)x-1 , x = 1, 2, …

    Find the probability distribution of Y = X2.

    Solution:

    – Since X ≥ 0, the transformation is one-to-one.

    – The inverse transform function is X = .

    – fY(y) = p(1-p)-1 , y = 1, 4, 9, 16,…

    Sec 5-5 General Functions of Random Variables 54

    y

    y

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    General Function of a Continuous Random Variable

    Suppose that X is a continuous random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability distribution of Y is

    fY(y) = fX[u(y)]∙|J|

    where J = u’(y) is called the Jacobian of the transformation and the absolute value of J is used.

    Sec 5-5 General Functions of Random Variables 55

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    Example 5-35: Function of a Continuous Random Variable

    Let X be a continuous random variable with probability

    distribution:

    Find the probability distribution of Y = h(X) = 2X + 4

    Sec 5-5 General Functions of Random Variables 56

    ( ) for 0 48

    X

    xf x x

    Note that has a one-to-one relationship to .

    4 1 and the Jacobian is '

    2 2

    4 2 1 4 for 4 12.

    8 2 32Y

    Y X

    yx u y J u y

    y yf y y

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    Definition of Moments about the Origin

    Sec 5-6 Moment Generating Functions 57

    The rth moment about the origin of the

    random variable X is

    ( ), discrete

    ' ( )( ) , continuous

    r

    r

    r r

    X f x X

    E XX f x dx X

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    Definition of a Moment-Generating Function

    Sec 5-6 Moment Generating Functions 58

    The moment-generating function of the random variable X is

    the expected value of etX and is denoted by MX (t). That is,

    ( ), discrete

    ( ) ( )( ) , continuous

    tX

    tX

    X tX

    e f x X

    M t M ee f x dx X

    Let X be a random variable with moment-generating

    function MX (t). Then

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-36 Moment-Generating Function for a Binomial Random Variable-1

    Sec 5-6 Moment Generating Functions 59

    Let X follows a binomial distribution, that is

    Determine the moment generating function and use it to verify that the

    mean and variance of the binomial random variable are μ=np and

    σ2=np(1-p).

    ( ) (1 ) , 0,1,....,x n xn

    f x p p x nx

    Now the first and second order derivatives will be

    0 0

    ( ) (1 ) ( ) (1 )n n

    tx x n x t x n x

    X

    x x

    n nM t e p p pe p

    x x

    The moment-generating function is

    which is the binomial expansion of

    [ (1 )]t npe p

    ' 1

    '' 2

    ( ) [1 ( 1)] and

    ( ) (1 )[1 ( 1)]

    t t n

    x

    t t t n

    x

    M t npe p e

    M t npe p npe p e

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Example 5-36 Moment-Generating Function for a Binomial Random Variable-2

    Sec 5-6 Moment Generating Functions 60

    If we set t = 0 in the above two equations we get

    ' '

    1

    '' '

    2

    ( ) and

    ( ) (1 )

    x

    x

    M t np

    M t np p np

    Now the variance is

    2 ' 2 2

    2

    2

    (1 ) ( )

    (1 )

    np p np np

    np np

    np p

    2Hence, the mean is and variance is (1 ).np np p

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Properties of Moment-Generating Function

    Sec 5-6 Moment Generating Functions 61

    1 2

    1 2

    1. ( ) ( )

    2. ( ) ( )

    If , ,..., are independent random vari

    If is a random variable

    ables with

    moment generat

    and is a co

    ing functions ( ), ( ),...,

    nstant, then

    ( )

    respectiv

    n

    at

    X a X

    aX X

    n

    X X X

    M t e M t

    M t M at

    X X X

    M t M t M t

    X a

    1 2

    1 2ely, and if ... then the moment

    generating function of Y is

    3. ( ) ( ). ( ). ... . ( )n

    n

    Y X X X

    Y X X X

    M t M t M t M t

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    Example 5-38 Distribution of a Sum of Poisson Random Variables

    Sec 5-6 Moment Generating Functions 62

    Suppose that X1 and X2 are two independent Poisson random variables

    with parameters λ1 and λ2, respectively. Determine the probability

    distribution of Y = X1 + X2.

    Using , the moment-generating

    function of Y = X1 + X2 is

    ( 1)( )te

    XM t e

    The moment-generating function of a Poisson random variable with

    parameter λ is

    Hence for X1 and X2,1 2

    1 2

    ( 1) ( 1)( ) and ( )

    t te e

    X XM t e M t e

    1 2

    1 2

    1 2

    ( 1) ( 1)

    ( )( 1)

    ( ) ( ). ( )

    t t

    t

    Y X X

    e e

    e

    M t M t M t

    e e

    e

    1 2( ) ( ). ( ). ... . ( )

    nY X X XM t M t M t M t

  • Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

    Important Terms & Concepts for Chapter 5

    Bivariate distribution

    Bivariate normal distribution

    Conditional mean

    Conditional probability density

    function

    Conditional probability mass

    function

    Conditional variance

    Contour plots

    Correlation

    Covariance

    Error propagation

    General functions of random

    variables

    Independence

    Joint probability density function

    Joint probability mass function

    Linear functions of random

    variables

    Marginal probability distribution

    Multinomial distribution

    Reproductive property of the

    normal distribution

    Chapter 5 Summary 63